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pychu [463]
2 years ago
8

Consider the situation||: A child pulls a sled by a rope across the lawn at a constant speed. Of the forces listed, identify whi

ch act upon the sled.
Normal, Gravity, Applied, Friction, Tension, & Air Resistance
Physics
1 answer:
Mice21 [21]2 years ago
6 0

Answer:

Gravitational

Tension

Normal

Friction.

Explanation:

The forces acting on the sled are:

Tension: the tension from the rope, this is the force that "moves" the sled.

Friction: kinetic friction between the sled and the ground as the sled moves.

There are another two forces that also act on the sled, but that "has no effect"

Gravitational force: This force pulls the sled down, against the floor.

Normal force: This force "opposes" to the gravitational one, so they cancel each other.

These two forces cancel each other, so they have no direct impact on the movement of the sled. BUT, the friction force depends on the weight of the moving object, and the weight of the moving object depends on the gravitational force, so we need gravitational force in order to have friction force.

Then we can conclude that the forces acting on the sled are:

Gravitational

Tension

Normal

Friction.

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A 20.00 kg lead sphere is hanging from a hook by a thin, massless wire 2.80 m long and is free to swing in a complete circle. Tr
Tema [17]

The minimum initial speed of the dart so that the combination makes a complete circular loop after the collision is 58.5 m/s.

<h3>Minimum speed for the object not fall out of the circle</h3>

The minimum speed if given by tension in the wire;

T + mg = ma

T + mg = m(v²)/R

tension must be zero for the object not fall

0 + mg = mv²/R

v = √(Rg)

<h3>Final speed of the two mass after collision</h3>

Use the principle of conservation of energy

K.Ef  = K.Ei + P.E

¹/₂mvf² = ¹/₂mv² + mg(2R)

¹/₂vf² = ¹/₂v² + g(2R)

¹/₂vf² = ¹/₂(Rg) + g(2R)

vf² = Rg + 4Rg

vf² = 5Rg

vf = √(5Rg)

vf = √(5 x 2.8 x 9.8)

vf = 11.7 m/s

<h3>Initial speed of the dart</h3>

Apply principle of conservation of linear momentum for inelastic collision;

5v = vf(20 + 5)

5v = 11.7(25)

5v = 292.5

v = 58.5 m/s

Learn  more about linear momentum here: brainly.com/question/7538238

#SPJ1

3 0
2 years ago
I NEED HELP ASAP! PLSSSSSSSSSSSSSSSSSSS
OleMash [197]
I’m pretty sure

Part 1; C
Part 2; C

Not 100% sure tho :)
5 0
3 years ago
Four point charges are placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of
kvv77 [185]

Answer:V_{net}=4\frac{kQ}{a}    

Explanation:

Given

charge on each Particle is Q

Length of diagonal of the square is 2a

therefore distance between center and each charge is \frac{2a}{2}=a

Electric Potential of charged Particle is given by

For First Charge

V_1=\frac{kQ}{a}

V_2=\frac{kQ}{a}

V_3=\frac{kQ}{a}

V_4=\frac{kQ}{a}

total Electric Potential At center is given by

V_{net}=V_1+V_2+V_3+V_4

V_{net}=4\times \frac{kQ}{a}

V_{net}=4\frac{kQ}{a}            

7 0
3 years ago
A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m
kenny6666 [7]

Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, N - mg = F_c

N-mg=\frac{mv^2}{r}

Now substitute the given values, we get

N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}

N=15.68+46.08

N = 61.76  N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

5 0
3 years ago
Plz solve completely
tiny-mole [99]

Answer:

Explanation:

wouldnt it just be x=0 sorry if im

wrong

6 0
3 years ago
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